Let's solve **Question 1 step-by-step using the Student t-test**. --- ## **Step 1: Write the Data** | Set | Data | |------|----------------------------------------| | set1 | 36.4, 42.4, 31.9, 36.6, 37.4, 41.9, 33.1 | | set2 | 30.9, 42.6, 28.4, 33.5, 38.2, 40.1, 34.1 | --- ## **Step 2: Calculate the Means** \[ \bar{x}_1 = \frac{36.4 + 42.4 + 31.9 + 36.6 + 37.4 + 41.9 + 33.1}{7} = \frac{259.7}{7} = 37.1 \] \[ \bar{x}_2 = \frac{30.9 + 42.6 + 28.4 + 33.5 + 38.2 + 40.1 + 34.1}{7} = \frac{247.8}{7} = 35.4 \] --- ## **Step 3: Calculate the Sample Variances** Use the formula: \[ s^2 = \frac{\sum (x_i - \bar{x})^2}{n-1} \] ### For Set 1: - Find squared differences from the mean (37.1): - (36.4 - 37.1)² = .49 - (42.4 - 37.1)² = 28.09 - (31.9 - 37.1)² = 27.04 - (36.6 - 37.1)² = .25 - (37.4 - 37.1)² = .09 - (41.9 - 37.1)² = 23.04 - (33.1 - 37.1)² = 16.00 - Sum = .49 + 28.09 + 27.04 + .25 + .09 + 23.04 + 16.00 = 94.99 - Variance: \( s_1^2 = \frac{94.99}{6} = 15.83 \) ### For Set 2: - Find squared differences from the mean (35.4): - (30.9 - 35.4)² = 20.25 - (42.6 - 35.4)² = 51.84 - (28.4 - 35.4)² = 49.00 - (33.5 - 35.4)² = 3.61 - (38.2 - 35.4)² = 7.84 - (40.1 - 35.4)² = 22.09 - (34.1 - 35.4)² = 1.69 - Sum = 20.25 + 51.84 + 49.00 + 3.61 + 7.84 + 22.09 + 1.69 = 156.32 - Variance: \( s_2^2 = \frac{156.32}{6} = 26.05 \) --- ## **Step 4: Apply the t-test Formula** \[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \] Where \( n_1 = n_2 = 7 \). \[ t = \frac{37.1 - 35.4}{\sqrt{\frac{15.83}{7} + \frac{26.05}{7}}} \] Calculate denominator: \[ \frac{15.83}{7} = 2.26 \] \[ \frac{26.05}{7} = 3.72 \] \[ 2.26 + 3.72 = 5.98 \] \[ \sqrt{5.98} = 2.45 \] So, \[ t = \frac{1.7}{2.45} = .694 \] --- ## **Step 5: Degrees of Freedom (df)** \[ df = n_1 + n_2 - 2 = 7 + 7 - 2 = 12 \] --- ## **Step 6: Find the Critical t-value (from table) for 50% confidence, df = 12** - For 50% confidence (two-tailed), \(\alpha = .5\). - From the table, for \(df = 12\), at .5: \(t_{critical} = .695\) --- ## **Step 7: Compare t-value** - Calculated t = .694 - Critical t = .695 \[ |t| < t_{critical} \] --- ## **Conclusion** - Since the calculated t-value is **less than** the critical t-value, we **fail to reject the null hypothesis**. - **The two data sets are statistically the same at the 50% confidence interval.** --- ### **Final Answer** > **At the 50% confidence interval, the two data sets are statistically the same.** --- **Let me know if you need the solution for Question 2 as well!**

## **Question 2: Linear Regression Analysis Step-by-Step** ### **Step 1: Write the Data** | x | y | |----|-----| | 12 | 100 | | 53 | 202 | | 35 | 31 | | a | 389 | | 55 | 532 | *Note: The value for 'a' is unspecified in the provided data. Assuming 'a' is a placeholder, you will need to replace it with a numerical value to proceed.* ### **Step 2: Calculate the Means** Assuming 'a' is a known number, calculate the means: \[ \bar{x} = \frac{\sum x_i}{N}, \quad \bar{y} = \frac{\sum y_i}{N} \] Where \(N\) is the number of data points. ### **Step 3: Calculate the Slope (m) and Intercept (b)** Using the formulas: \[ m = \frac{N \sum (x_i y_i) - \sum x_i \sum y_i}{N \sum (x_i^2) - (\sum x_i)^2} \] \[ b = \bar{y} - m \bar{x} \] ### **Step 4: Compute Required Sums** 1. Calculate \( \sum x_i \), \( \sum y_i \), \( \sum (x_i y_i) \), and \( \sum (x_i^2) \). 2. Substitute values into the formulas for \(m\) and \(b\). ### **Step 5: Calculate the Correlation Coefficient (r)** Using the formula: \[ r = \frac{N \sum (x_i y_i) - \sum x_i \sum y_i}{\sqrt{[N \sum (x_i^2) - (\sum x_i)^2][N \sum (y_i^2) - (\sum y_i)^2]}} \] ### **Step 6: Calculate the Standard Error (SE)** Standard error of estimate is given by: \[ SE = \sqrt{\frac{\sum (y_i - \hat{y}_i)^2}{N-2}} \] Where \(\hat{y}_i\) are the predicted \(y\) values from the regression line. ### **Final Equations** - **Regression Line:** \(y = mx + b\) - **Correlation Coefficient:** \(r\) - **Standard Error:** \(SE\) ### **Conclusion** 1. **Regression Equation:** Replace \(m\) and \(b\) with computed values. 2. **Correlation Coefficient:** Replace \(r\) with computed value. 3. **Standard Error:** Replace \(SE\) with computed value. ### **Final Answer** > The linear regression equation, correlation coefficient, and standard error will be derived after calculating the necessary sums using the complete dataset. Ensure 'a' is defined for accurate computation. --- If you provide a specific value for 'a', I can help finalize the calculations!

## **Question 2: Linear Regression Analysis** ### **Given Data** Assuming 'a' is 45 for computation: | x | y | |----|-----| | 12 | 100 | | 53 | 202 | | 35 | 31 | | 45 | 389 | | 55 | 532 | ### **Step 1: Calculate Sums** 1. **Number of Data Points (N):** 5 2. **Sums:** - \( \sum x = 12 + 53 + 35 + 45 + 55 = 200 \) - \( \sum y = 100 + 202 + 31 + 389 + 532 = 1254 \) - \( \sum (x \cdot y) = 12 \cdot 100 + 53 \cdot 202 + 35 \cdot 31 + 45 \cdot 389 + 55 \cdot 532 = 1200 + 10706 + 1085 + 17505 + 29260 = 62256 \) - \( \sum (x^2) = 12^2 + 53^2 + 35^2 + 45^2 + 55^2 = 144 + 2809 + 1225 + 2025 + 3025 = 10228 \) ### **Step 2: Calculate the Slope (m) and Intercept (b)** #### **Slope (m) Formula:** \[ m = \frac{N \sum (x_i y_i) - \sum x \sum y}{N \sum (x_i^2) - (\sum x)^2} \] Substituting the values: \[ m = \frac{5 \cdot 62256 - 200 \cdot 1254}{5 \cdot 10228 - 200^2} \] Calculating: \[ m = \frac{311280 - 250800}{51140 - 40000} = \frac{60480}{11140} \approx 5.43 \] #### **Intercept (b) Formula:** \[ b = \bar{y} - m \bar{x} \] Where: \[ \bar{x} = \frac{\sum x}{N} = \frac{200}{5} = 40 \] \[ \bar{y} = \frac{\sum y}{N} = \frac{1254}{5} = 250.8 \] Calculating: \[ b = 250.8 - 5.43 \cdot 40 \approx 250.8 - 217.2 \approx 33.6 \] ### **Step 3: Regression Equation** The linear regression equation is: \[ y = 5.43x + 33.6 \] ### **Step 4: Correlation Coefficient (r)** Using the formula: \[ r = \frac{N \sum (x_i y_i) - \sum x \sum y}{\sqrt{[N \sum (x_i^2) - (\sum x)^2][N \sum (y_i^2) - (\sum y)^2]}} \] Calculating \( \sum (y^2) \): \[ \sum (y^2) = 100^2 + 202^2 + 31^2 + 389^2 + 532^2 = 10000 + 40804 + 961 + 152921 + 283024 = 476710 \] Substituting into the \(r\) formula: \[ r = \frac{5 \cdot 62256 - 200 \cdot 1254}{\sqrt{(5 \cdot 10228 - 200^2)(5 \cdot 476710 - 1254^2)}} \] Calculating: \[ r = \frac{311280 - 250800}{\sqrt{(51140 - 40000)(2383550 - 1570516)}} \] \[ = \frac{60480}{\sqrt{11140 \cdot 813034}} \approx \frac{60480}{\sqrt{9070901960}} \approx \frac{60480}{3009.15} \approx 0.20 \] ### **Step 5: Standard Error (SE)** Calculating predicted \(y\) values \( \hat{y} \) using the regression equation for each \(x\): 1. For \(x = 12\): \( \hat{y} = 5.43 \cdot 12 + 33.6 \approx 94.76 \) 2. For \(x = 53\): \( \hat{y} = 5.43 \cdot 53 + 33.6 \approx 333.79 \) 3. For \(x = 35\): \( \hat{y} = 5.43 \cdot 35 + 33.6 \approx 205.85 \) 4. For \(x = 45\): \( \hat{y} = 5.43 \cdot 45 + 33.6 \approx 265.41 \) 5. For \(x = 55\): \( \hat{y} = 5.43 \cdot 55 + 33.6 \approx 324.96 \) Calculating residuals \( (y - \hat{y})^2 \): 1. \( (100 - 94.76)^2 \approx 26.25 \) 2. \( (202 - 333.79)^2 \approx 17680.34 \) 3. \( (31 - 205.85)^2 \approx 30283.82 \) 4. \( (389 - 265.41)^2 \approx 15159.39 \) 5. \( (532 - 324.96)^2 \approx 43340.48 \) Sum of squared residuals: \[ \sum (y - \hat{y})^2 \approx 26.25 + 17680.34 + 30283.82 + 15159.39 + 43340.48 \approx 106490.28 \] Calculating SE: \[ SE = \sqrt{\frac{\sum (y - \hat{y})^2}{N-2}} = \sqrt{\frac{106490.28}{3}} \approx 188.74 \] ### **Final Answers** - **Regression Equation:** \(y = 5.43x + 33.6\) - **Correlation Coefficient (r):** \(0.20\) - **Standard Error (SE):** \(188.74\) If you have further questions or need adjustments, please let me know!

## **Question 2: Linear Regression Analysis Using Least Squares Method** ### **Given Data** | x | y | |------|--------| | 1 | -0.96 | | 2 | -1.94 | | 3 | -2.92 | | 3.5 | -3.41 | | 4 | -3.89 | | 5.5 | -5.32 | ### **Step 1: Calculate Sums** 1. **Number of Data Points (N):** 6 2. **Sums:** - \( \sum x = 1 + 2 + 3 + 3.5 + 4 + 5.5 = 19 \) - \( \sum y = -0.96 - 1.94 - 2.92 - 3.41 - 3.89 - 5.32 = -18.44 \) - \( \sum (x \cdot y) = 1 \cdot (-0.96) + 2 \cdot (-1.94) + 3 \cdot (-2.92) + 3.5 \cdot (-3.41) + 4 \cdot (-3.89) + 5.5 \cdot (-5.32) \) \[ \sum (x \cdot y) = -0.96 - 3.88 - 8.76 - 11.94 - 15.56 - 29.36 = -70.46 \] - \( \sum (x^2) = 1^2 + 2^2 + 3^2 + 3.5^2 + 4^2 + 5.5^2 \) \[ \sum (x^2) = 1 + 4 + 9 + 12.25 + 16 + 30.25 = 72.5 \] ### **Step 2: Calculate the Slope (a1) and Intercept (a0)** #### **Slope (a1) Formula:** \[ a_1 = \frac{N \sum (x \cdot y) - \sum x \sum y}{N \sum (x^2) - (\sum x)^2} \] Substituting the values: \[ a_1 = \frac{6 \cdot (-70.46) - 19 \cdot (-18.44)}{6 \cdot 72.5 - 19^2} \] Calculating: \[ a_1 = \frac{-422.76 + 350.36}{435 - 361} = \frac{-72.4}{74} \approx -0.978 \] #### **Intercept (a0) Formula:** \[ a_0 = \bar{y} - a_1 \bar{x} \] Where: \[ \bar{x} = \frac{\sum x}{N} = \frac{19}{6} \approx 3.167 \] \[ \bar{y} = \frac{\sum y}{N} = \frac{-18.44}{6} \approx -3.0733 \] Calculating \(a_0\): \[ a_0 = -3.0733 - (-0.978) \cdot 3.167 \approx -3.0733 + 3.096 \approx 0.0227 \] ### **Step 3: Regression Equation** The linear regression equation is: \[ y = 0.0227 - 0.978x \] ### **Step 4: Correlation Coefficient (r)** Using the formula: \[ r = \frac{N \sum (x \cdot y) - \sum x \sum y}{\sqrt{[N \sum (x^2) - (\sum x)^2][N \sum (y^2) - (\sum y)^2]}} \] Calculating \( \sum (y^2) \): \[ \sum (y^2) = (-0.96)^2 + (-1.94)^2 + (-2.92)^2 + (-3.41)^2 + (-3.89)^2 + (-5.32)^2 \] Calculating each term: \[ = 0.9216 + 3.7616 + 8.5364 + 11.6281 + 15.1521 + 28.3024 = 68.3022 \] Substituting into the \(r\) formula: \[ r = \frac{6 \cdot (-70.46) - 19 \cdot (-18.44)}{\sqrt{(6 \cdot 72.5 - 19^2)(6 \cdot 68.3022 - (-18.44)^2)}} \] Calculating: \[ = \frac{-422.76 + 350.36}{\sqrt{(435 - 361)(409.8132 - 339.6336)}} \] \[ = \frac{-72.4}{\sqrt{74 \cdot 70.1796}} = \frac{-72.4}{\sqrt{5193.2744}} \approx \frac{-72.4}{72.1} \approx -1.004 \] ### **Step 5: Standard Error (SE)** Calculating predicted \(y\) values \( \hat{y} \): 1. For \(x = 1\): \( \hat{y} = 0.0227 - 0.978 \cdot 1 \approx -0.9553 \) 2. For \(x = 2\): \( \hat{y} = 0.0227 - 0.978 \cdot 2 \approx -1.9333 \) 3. For \(x = 3\): \( \hat{y} = 0.0227 - 0.978 \cdot 3 \approx -2.9113 \) 4. For \(x = 3.5\): \( \hat{y} = 0.0227 - 0.978 \cdot 3.5 \approx -3.3988 \) 5. For \(x = 4\): \( \hat{y} = 0.0227 - 0.978 \cdot 4 \approx -4.3768 \) 6. For \(x = 5.5\): \( \hat{y} = 0.0227 - 0.978 \cdot 5.5 \approx -5.3548 \) Calculating residuals \( (y - \hat{y})^2 \): 1. \( (-0.96 - -0.9553)^2 \approx 0.000021 \) 2. \( (-1.94 - -1.9333)^2 \approx 0.000045 \) 3. \( (-2.92 - -2.9113)^2 \approx 0.000074 \) 4. \( (-3.41 - -3.3988)^2 \approx 0.000129 \) 5. \( (-3.89 - -4.3768)^2 \approx 0.237969 \) 6. \( (-5.32 - -5.3548)^2 \approx 0.001235 \) Sum of squared residuals: \[ \sum (y - \hat{y})^2 \approx 0.000021 + 0.000045 + 0.000074 + 0.000129 + 0.237969 + 0.001235 \approx 0.239473 \] Calculating SE: \[ SE = \sqrt{\frac{\sum (y - \hat{y})^2}{N-2}} = \sqrt{\frac{0.239473}{4}} \approx 0.245 \] ### **Final Answers** - **Regression Equation:** \(y = 0.0227 - 0.978x\) - **Correlation Coefficient (r):** \(r \approx -1.00\) (perfect negative correlation) - **Standard Error (SE):** \(SE \approx 0.245\) If you have any further questions or need additional details, please let me know!